The given consists of of . When using the to solve a system of equations, it is necessary to isolate a . In the third equation,
z is already isolated, so we will start by substituting its value into the second equation.
⎩⎪⎪⎨⎪⎪⎧13=3x−y4y−3x+2z=-3z=2x−4y
⎩⎪⎪⎨⎪⎪⎧13=3x−y4y−3x+2(2x−4y)=-3z=2x−4y
⎩⎪⎪⎨⎪⎪⎧13=3x−y4y−3x+4x−8y=-3z=2x−4y
⎩⎪⎪⎨⎪⎪⎧13=3x−yx−4y=-3z=2x−4y
⎩⎪⎪⎨⎪⎪⎧13=3x−yx=4y−3z=2x−4y
This time, the
x-variable was isolated in the second equation. We can now substitute its into the first equation.
⎩⎪⎪⎨⎪⎪⎧13=3x−yx=4y−3z=2x−4y
⎩⎪⎪⎨⎪⎪⎧13=3(4y−3)−yx=4y−3z=2x−4y
⎩⎪⎪⎨⎪⎪⎧13=12y−9−yx=4y−3z=2x−4y
⎩⎪⎪⎨⎪⎪⎧13=11y−9x=4y−3z=2x−4y
⎩⎪⎪⎨⎪⎪⎧22=11yx=4y−3z=2x−4y
⎩⎪⎪⎨⎪⎪⎧2=yx=4y−3z=2x−4y
⎩⎪⎪⎨⎪⎪⎧y=2x=4y−3z=2x−4y
The value of
y is
2. Substituting
2 for
y into the second equation, we can find the value of
x.
⎩⎪⎪⎨⎪⎪⎧y=2x=4y−3z=2x−4y
⎩⎪⎪⎨⎪⎪⎧y=2x=4(2)−3z=2x−4y
⎩⎪⎪⎨⎪⎪⎧y=2x=8−3z=2x−4y
⎩⎪⎪⎨⎪⎪⎧y=2x=5z=2x−4y
Now that we know the values of
x and
y, we are able to find the value of
z.
⎩⎪⎪⎨⎪⎪⎧y=2x=5z=2x−4y
⎩⎪⎪⎨⎪⎪⎧y=2x=5z=2(5)−4(2)
▼
(III): Evaluate right-hand side
⎩⎪⎪⎨⎪⎪⎧y=2x=5z=2
The solution to the system is the point
(5,2,2). This is the singular point at which all three planes . Let's check our solution by substituting the values into the system.
⎩⎪⎪⎨⎪⎪⎧13=3x−y4y−3x+2z=-3z=2x−4y
(I), (II), (III): Substitute values
⎩⎪⎪⎪⎨⎪⎪⎪⎧13=?3(5)−24(2)−3(5)+2(2)=?-32=?2(5)−4(2)
(I), (II), (III): Multiply
⎩⎪⎪⎪⎨⎪⎪⎪⎧13=?15−28−15+4=?-32=?10−8
(I), (II), (III): Add and subtract terms
⎩⎪⎪⎨⎪⎪⎧13=13 ✓-3=-3 ✓2=2 ✓
Since the substitution of our answers into the given equations resulted in three , we know that our solution is correct.